How do you convert cartesian $(- \sqrt{7}, 0)$ $(-sqrt7, 0)$ to polar form?

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How do you convert cartesian $(- \sqrt{7}, 0)$ $(-sqrt7, 0)$ to polar form?

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Answer 1 · 2016-04-27T14:42:04

Cartesian form $(x, y) = (- \sqrt{7}, 0) = (r cos θ, r sin θ)$ $(x, y) = (-sqrt 7, 0) = (r cos theta, r sin theta)$ transforms to the polar form $(r, θ) = (\sqrt{7}, π)$ $(r, theta ) = ( sqrt 7, pi )$ .

Explanation:

$x = - \sqrt{7} and y = 0$ $x=-sqrt 7 and y=0$ .

$r = \sqrt{x^{2} + y^{2}} = \sqrt{7} . cos θ = \frac{x}{r} = - 1 and sin θ = \frac{y}{r} = 0$ $r=sqrt(x^2+y^2)=sqrt 7. cos theta = x/r=-1 and sin theta =y/r=0$
So, $θ = π$ $theta=pi$ ..

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How do you convert cartesian $(- \sqrt{7}, 0)$ $(-sqrt7, 0)$ to polar form?

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How do you convert cartesian (−√7,0)(-sqrt7, 0) to polar form?

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How do you convert cartesian $(- \sqrt{7}, 0)$ $(-sqrt7, 0)$ to polar form?